Question

Consider a rod of length 30 for which \(\alpha^{2}=1 .\) Suppose the initial temperature distribution is given by \(u(x, 0)=x(60-x) / 30\) and that the boundary conditions are \(u(0, t)=30\) and \(u(30, t)=0\) (a) Find the temperature in the rod as a function of position and time. (b) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x\). (c) Plot \(u\) versus \(t\) for \(x=12\). Observe that \(u\) initially decreases, then increases for a while, and finally decreases to approach its steady-state value. Explain physically why this behavior occurs at this point.

Short answer

Answer: The behavior of the temperature at \(x=12\) shows an initial decrease, followed by an increase, and then a final decrease approaching the steady-state value. The physical interpretation of this behavior is that the externally supplied temperature is absorbed at the specific point in the rod, spreading the heat along the rod through conduction, causing the local temperature to increase for a brief period. As the heat is evenly distributed along the entire rod, the local temperature starts to decrease and approaches the steady-state temperature distribution.

Step-by-step solution

Set up the heat equation

The heat equation is: $$\frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2}$$ where \(\alpha = 1\), so: $$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$

Apply separation of variables

Let \(u(x,t) = X(x)T(t)\). Then we have: $$X(x)\frac{dT(t)}{dt} = \frac{d^2 X(x)}{dx^2}T(t)$$ Now divide both sides by \(X(x)T(t)\): $$\frac{1}{T(t)}\frac{dT(t)}{dt} = \frac{1}{X(x)}\frac{d^2 X(x)}{dx^2} = -\lambda$$

Solve for X(x)

We have: $$\frac{d^2 X(x)}{dx^2} = -\lambda X(x)$$ Solve this second order ODE for \(X(x)\). The solution will have the form: $$X_n(x) = A_n \sin\left(\sqrt{\lambda_n}x\right) + B_n \cos\left(\sqrt{\lambda_n}x\right)$$ We apply the boundary conditions to find the coefficients \(A_n\) and \(B_n\): $$X(0) = 0 \implies B_n = 0$$ $$X(30) = 0 \implies \lambda_n = \frac{(n\pi)^2}{30^2}$$ The final function for \(X_n(x)\) becomes: $$X_n(x) = A_n \sin\left(\frac{n\pi x}{30}\right)$$

Solve for T(t)

We have: $$\frac{dT(t)}{dt} = -\lambda T(t)$$ Solve this first order ODE for \(T(t)\). The solution will have the form: $$T_n(t) = C_n e^{-\lambda_n t}$$ where \(C_n\) is another set of coefficients.

Apply the initial condition

Now, we apply the initial condition \(u(x,0) = x(60-x) / 30\) to find the coefficients \(A_n\) and \(C_n\): $$\frac{x(60-x)}{30} = \sum_{n=1}^{\infty} A_n C_n \sin\left(\frac{n\pi x}{30}\right)$$ Using Fourier sine series expansion, we find the coefficients \(A_n\): $$A_n = \frac{2}{30} \int_0^{30} x(60-x) \sin\left(\frac{n\pi x}{30}\right) dx$$ Solve this integral and plug \(A_n\) back into the \(X_n(x)\) expression. Since \(T(0)=1,\) we can conclude that \(C_n=1\) for every \(n.\) Thus, the final solution for \(u(x,t)\) is: $$u(x,t) = \sum_{n=1}^{\infty}A_n \sin\left(\frac{n\pi x}{30}\right) e^{-\lambda_n t}$$

Plotting and analysis

We can use a software like MATLAB or Python's Matplotlib to plot the function \(u(x,t)\) for various values of \(x\) and \(t\) according to the problem statement. For part (c), we want to plot \(u\) versus \(t\) for \(x=12\). We will observe the behavior of the temperature initially decreasing, increasing for a while, and then finally decreasing to approach its steady-state value. The physics behind this behavior could be that the externally supplied temperature is absorbed at the specific point in the rod. Due to conduction, the heat will then spread along the rod, increasing the temperature for a while. Once the heat is evenly distributed along the entire rod, the local temperature will start to decrease as it approaches the steady-state temperature distribution.

Key concepts

Partial Differential Equations

Imagine you have a rod and you want to model how heat flows through it over time. This is where partial differential equations (PDEs) come into play. PDEs are equations involving rates of change with respect to more than one variable. In the context of our heat equation, we have variables for both space and time.
The heat equation itself is a classic example of a PDE. It's written as:
\[ \frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2} \]
Here, \(u(x,t)\) is the temperature at position \(x\) and time \(t\). \(\alpha^2\) is a constant related to the material's thermal properties. PDEs help describe how various physical processes like heat, sound, and fluid dynamics behave spatially and temporally. Understanding PDEs is vital for solving real-world problems, especially where changes occur continuously over space and time.

Boundary Value Problems

In many real-world scenarios, not only do you need to solve the equations, but you also have to respect certain conditions at the edges or boundaries of your domain. These are called boundary value problems. In our rod example, the boundaries are at \(x=0\) and \(x=30\). They define how the rod interacts with its surroundings.
The given conditions:

• \(u(0, t) = 30\)
• \(u(30, t) = 0\)

specify the temperatures at the two ends of the rod. These conditions ensure that the solution is unique and physically meaningful.
When working with boundary value problems, one typically encounters either Dirichlet boundaries, where the function's value is specified, or Neumann boundaries, where the function's derivative is specified. By incorporating these conditions, we can make predictions about how heat will behave across the entire length of the rod.

Separation of Variables

To tackle partial differential equations effectively, a handy method is called separation of variables. This method assumes the solution can be broken down into functions of each variable separately. For the heat equation, let's say \(u(x, t) = X(x) T(t)\).
By substituting, we break the PDE into two ordinary differential equations (ODEs), each dependent on a single variable.

• \(\frac{1}{T(t)}\frac{dT(t)}{dt} = -\lambda\)
• \(\frac{1}{X(x)}\frac{d^2 X(x)}{dx^2} = -\lambda\)

This approach simplifies solving the original equation by reducing it to simpler ODEs. It's particularly useful because it transforms a complex, multi-variable problem into more manageable pieces.
Separation of variables often works well for problems with boundary and initial conditions, like in our rod scenario, providing a path to a physical solution by finding each function and combining them back together.

Fourier Series

When you have solutions expressed in infinite sums of sines and cosines, you're dealing with a Fourier series. Fourier series are powerful tools to handle periodic functions or signals akin to them.
In our heat equation, we used Fourier sine series to express the initial condition and find the coefficients. The series takes the form:
\[ u(x, t) = \sum_{n=1}^{\infty}A_n \sin\left(\frac{n\pi x}{30}\right) e^{-\lambda_n t} \]
Where \(A_n\) are coefficients found using integrals derived from the initial condition.
A Fourier series allows us to approximate more complex functions by adding together simple waves. In the case of heat distribution along the rod, it lets us break down the problem into manageable parts.
Using Fourier series techniques, even challenging initial conditions can be efficiently handled, transforming them into opportunities to apply elegant mathematical treatments to physical phenomena.